SOLUTION: A person has 11 coins consisting of dimes and nickels. If the total amount of money is $0.75, how much of each coin are there?

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Question 1170525: A person has 11 coins consisting of dimes and nickels. If the total amount of money is $0.75, how much of each coin are there?
Found 3 solutions by josgarithmetic, greenestamps, ikleyn:
Answer by josgarithmetic(39620)   (Show Source): You can put this solution on YOUR website!
Since you have only eleven coins for seventy five cents, you could simply look at several of the possible combinations of nickels and dimes and find the combination that works.

You could use a little algebra instead.
d dimes
11-d nickels

Simplify and solve.

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


A quick informal way to find the solution, without formal algebra, but also without blind trial and error....

(1) Imagine you have 11 coins that are all nickels; the total value would be 55 cents.
(2) The actual total value is 75 cents, which is 20 cents more than that.
(3) Exchange nickels for dimes one at a time, thus keeping the number of coins the same but increasing the total value by 5 cents each time.
Since you need to increase the value by a total of 20 cents, you need to do that exchange 20/5 = 4 times. So you end up with 4 dimes and 11-4=7 nickels.

ANSWER: 4 dimes and 7 nickels

CHECK: 4(10)+7(5) = 40+35 = 75


Answer by ikleyn(52799)   (Show Source): You can put this solution on YOUR website!
.

Let d be the number of dimes; then (11-d) is the number of nickels.


Next, you write the total money equation


    10d + 5*(11-d) = 75   cents.


You cancel factor of 5 in both sides


     2d + (11-d) = 15


and simplify


    2d - d = 15 - 11,   or

       d   = 4.


ANSWER.  4 dimes and  11-4 = 7 nickels.

Solved.

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On coin problems,  see the lessons
    - Coin problems
    - More Coin problems
    - Solving coin problems without using equations
    - Kevin and Randy Muise have a jar containing coins
    - Typical coin problems from the archive
    - Three methods for solving standard (typical) coin word problems
    - More complicated coin problems
    - Advanced coin problems
    - Solving coin problems mentally by grouping without using equations
    - Non-typical coin problems
    - Santa Claus helps solving coin problem
    - OVERVIEW of lessons on coin word problems
in this site.

You will find there the lessons for all levels - from introductory to advanced,
and for all methods used - from one equation to two equations and even without equations.

A convenient place to quickly observe these lessons from a  "bird flight height"  (a top view)  is the last lesson in the list.

Read them and become an expert in solution of coin problems.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Coin problems".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.



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